Maps That Take Lines to Circles, in Dimension 4

Home , Hopf fibration

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Theorem 1 Suppose that an analytic diﬀeomorphism f between an open set in RP4 and an open set in S4 takes all lines to circles. Then it is either a restriction of a classical projection or a restriction of a (left or right) quaternionic Hopf ﬁbration.

Remark. When we say that a map f : U ⊆ RP4 → S4 is a restriction of a classical projection, we mean the following. There is a projective map i from RP4 to R5 defined everywhere on U, a conformal identification j of S4 with a Euclidean sphere in R5 and a central projection π to this Euclidean sphere from some point such that f = j−1 ◦ π ◦ i. Analogously, the statement “f is a restriction of a quaternionic Hopf fibration” has the following meaning. There is a projective map i form RP4 to RP7, a (right or left) quaternionic Hopf fibration π : RP7 → S4 and a conformal identification j of the sphere in the image of f with the sphere in the image of π such that f = j−1 ◦ π ◦ i. Note in particular, that Theorem 1 is insensitive to a projective transformation in the preimage and a conformal transformation in the image. Actually, it is not hard to see that we can always assume j to be fixed from the very beginning.

History of the problem. The problem of finding maps that take lines to circles came from nomography (for an introduction to nomography see [2]). G.S. Khovanskii in 1970-s posed the following question: find all diffeomorphisms between open subsets of R2 that take lines to circles, or, in the language of nomography, that transform nomograms with aligned points to circular nomograms. Circular nomograms are more convenient in practice whereas nomograms with aligned points are theoretically easier to deal with. This problem was solved by A.G. Khovanskii [3] who proved that all such diffeomorphisms come from projections of planes in R3 to Euclidean spheres. Actually, his result was stated in a different form: up to a projective transformation in the preimage and a Möbius transformation in the image there are only 3 such diffeomorphisms, and they correspond to classical geometries. Izadi in [4] extended the results of Khovanskii to the 3-dimensional case. It turned out (see [5]) that Khovanskii’s theorem does not extend to dimension 4. The simplest counterexample is a complex projective transformation which takes lines to circles but does not come from a projection of a hyperplane to a sphere. This is a particular case of example 2 above. In [5] it was proved that all ample enough rectifiable bundles of circles passing through a point in R4 are obtained by means of example 2 (those obtained from example 1 can be obtained from example 2 as well). This may be regarded as a first step in proving Theorem 1. This article completes the proof.

Outline of the proof of Theorem 1. Suppose that f : U → V is a diffeomorphism taking all lines to circles. First choose a projective identification of U with a region in R4 and a conformal identification of V with a region in H, the skew-field of quaternions. Then f can be regarded as a quaternion-valued function on U. For any constant vector field α denote by ∂α the Lie derivative along α. Put Aα = ∂αf. The quaternions Aα are best to be understood as components of a quaternion-

2 valued differential 1-form A on U, namely, the differential of f. This form is closed: dA = 0 or, in components, ∂αAβ = ∂βAα for any pair of constant vector fields α and β on U. By Lemma 5.2 from [5], at every point of U there exists a quaternion Bα that depends linearly on α and satisfies one of the equations

∂αAα = BαAα or ∂αAα = AαBα.

The products in the right-hand sides are in the sense of quaternions. Assume that the first equation holds everywhere on U (this is not very restrictive: see the end of Section 4). We show in Section 2 that the integrability condition for this equation is 1 1 ∂ B = B B + C(A A + A A + A A ). α β 2 α β 3 α β β α α β where C is a quaternion-valued function. We call it the first integrability condition. Then we use once again that f takes lines to circles, now to deduce that C is real- valued. If C = 0 identically, then the first integrability condition is integrable, and it gives us the maps from example 2. This will be proved in Section 3. Suppose now that C 6= 0. Then we need to write down an integrability condition for the first integrability condition — the second integrability condition. It will be obtained in Section 4. The second integrability condition turns out to be a differential relation between A, B and C only, with no additional parameters. It expresses all first logarithmic derivatives of C in terms of the values of A and B. The fact that C is real imposes a restriction on possible values of A and B. Namely, Bα = p(Aα)+ Aαq where p is a real-valued 1-form and q is a quaternion. Now fix some admissible values of A, B and C at some point of U. We will show in Section 5 that there exists a map from example 1 with the same values of A, B and C at the given point. On the other hand, from the second integrability condition it follows that such analytic map is unique. Thus our map f is from example 1.

2 First integrability condition

In this section, we obtain an integrability condition for the equation

∂αAα = BαAα. (1)

Since Aα are derivatives, we have ∂αAβ = ∂βAα. From this and from equation (1) it follows immediately that 1 ∂ A = (B A + B A ). (2) α β 2 α β β α

An integrability condition for equation (1) is obtained from the equality ∂α∂βAγ = ∂β∂αAγ by expressing derivatives of A with the help of (2). If we introduce a tensor C given in components by 1 C = ∂ B − B B , αβ α β 2 α β

3 then the integrability condition reads as follows:

Cαβ Aγ + Cαγ Aβ = CβαAγ + Cβγ Aα. (3)

If we apply to equation (3) the transposition of indices β ↔ γ, then we have:

′ Cαβ Aγ + Cαγ Aβ = CγαAβ + CγβAα. (3 )

Take the sum of equations (3) and (3′). It can be written as follows:

′′ (Cβγ + Cγβ)Aα = (2Cαγ − Cγα)Aβ + (2Cαβ − Cβα)Aγ . (3 )

Put β = γ in equation (3). We obtain that

(2Cαβ − Cβα)Aβ = CββAα. (4) or, replacing β with γ, ′ (2Cαγ − Cγα)Aγ = CγγAα. (4 ) ′′ Replace 2Cαβ − Cβα and 2Cαγ − Cγα in (3 ) by their expressions obtained from (4) and (4′), respectively. We obtain

−1 −1 −1 −1 Cβγ + Cγβ = CββAα(Aβ Aγ )Aα + CγγAα(Aγ Aβ )Aα . (5) Take a point in U. In equation (5) written at this point, α, β and γ may be arbitrary vectors. Fix β and γ, and let α vary. Then Aα runs over all quaternions, since f is a diﬀeomorphism. We need the following

Lemma 2 Let a and b be arbitrary quaternions, and x an imaginary quaternion (i.e. not a real number). Suppose that the number ay + by−1 stays the same for all quaternions y obtained from x by inner conjugations (i.e. y = qxq−1 for an arbitrary quaternion q). Then b = a|x|2.

Proof. Denote by x0 the real part of x. Then it is easy to see that

− b 2bx0 ay + by 1 = a − y + . |x|2 |x|2 The second term in the right-hand side is independent of y. Therefore, the ﬁrst term should be also independent of y which is possible only if the coeﬃcient a − b/|x|2 vanishes. We can apply this lemma to equation (5) where we put a = Cββ, b = Cγγ and −1 x = Aβ Aγ . If β is not parallel to γ, then x is imaginary. Thus by the lemma we have

Cββ Cγγ 2 = 2 . |Aβ| |Aγ | It follows that the left-hand side is independent of β. Hence

2 Cββ = C|Aβ| (6)

4 where C is a quaternion which does not depend on the vector β, but it may depend on a point — so C is a quaternion-valued function on U. Plug in (6) to (5): ′ Cβγ + Cγβ = C(Aβ Aγ + Aγ Aβ). (6 ) Comparing equations (6′) (with β changed to α and γ changed to β) and (4) we can conclude that 1 C = C(A A + A A + A A ). (6′′) αβ 3 α β β α α β This is the ﬁrst integrability condition.

3 Examples

Let us now take a closer look on examples 1 and 2 form Section 1.

Example 1. Consider the left quaternionic Hopf fibration π : RP7 → S4. A point in the preimage is represented by a pair of quaternions (y,z) up to multiplication of both y and z by a common real factor. The image of this point under π is an element of the left projective quaternionic line with homogeneous coordinates y and z. We can take y−1z as an affine conformal quaternionic coordinate in the image. Suppose that f is obtained as the composition of π with some projective embedding of (a part of) R4 to RP7. This projective embedding sends a point x ∈ R4 to a point in RP7 with coordinates y = L(x), z = M(x), where L and M are some affine maps from R4 to R4. Thus in the given coordinates f looks as follows: −1 f : x 7→ L(x) M(x) (E1) where the multiplication and the inverse are in the sense of quaternions. Denote by L~ and M~ the linear parts of L and M, respectively. This means that L~ (resp., M~ ) is a linear operator such that L (resp., M) is a composition of L~ (resp., M~ ) and a translation. In other words, L~ and M~ are differentials of L and M at any point. Compute Aα = ∂αf for our map f:

−1 −1 −1 Aα(x)= −L (x)L~ (α)L (x)M(x)+ L (x)M~ (α).

Diﬀerentiating once again along α we obtain:

−1 Bα = −2L (x)L~ (α).

1 We can see from here that ∂αBβ = 2 BαBβ, so in this case C = 0.

5 Case C = 0. Let us work out the case when C vanishes everywhere in U. In this case the first integrability condition (6′′) reads as follows: 1 ∂ B = B B . α β 2 α β Consider this equation together with equation (2). This is a system of partial differential equations of first order which expresses all first derivatives of A and B through the values of A and B. Denote this system by S. Fix any point x ∈ U and initial values of 1-forms A and B at x. Thus for any vector α at x we know Aα and Bα. Note that if a solution of S with given initial values exists, then it is unique. Indeed, all higher derivatives of A and B can be expressed through the initial values by means of the system S. The existence of a solution follows from example 1. In this example, at the given point x linear maps A : α 7→ Aα and B : α 7→ Bα can be arbitrary. We have just proved the following Proposition 3 If C = 0 identically in U, then the general solution of equations (2) ′′ and (6 ) is given by (E1).

Example 2. Consider a Euclidean sphere S in R5. We can choose this sphere to be centered at the origin and to have radius 1. Introduce a conformal coordinate system on S by means of the stereographic projection j to the equatorial hyperplane with the center at the Noth pole. To this end we need to choose an orthogonal splitting R5 = R4 × R the latter factor being the line between the North and South poles. Thus any point in R5 is represented as (y,z) where y ∈ R4 and z ∈ R. The point on S corresponding to (y, 0) by means of the stereographic projection is 2 − 2y |y| − 1 j 1(y)= , . 1+ |y|2 1+ |y|2

Suppose that f is the central projection from a horizontal hyperplane z = z1 to S with the center (0,z0). Under this projection, a point (x, z1) corresponds to a point j−1(y) on S if and only if

2y(z0 − z1) x = 2 . (z0 +1)+(z0 − 1)|y|

We can now rescale y and choose specific values of z0 and z1 so that y x = . (E2.1) 1+ |y|2 This can be considered as an implicit equation defining f : x 7→ y. Recall that f was defined as a classical projection. Another interpretation of f is that it estab- lishes a correspondence between the Klein and the Poincarémodels of the hyperbolic (Lobachevsky) geometry. 2 Multiply both sides of equation (E2.1) by the denominator to get y = (1+ |y| )x and then differentiate both sides along α:

2hy, Aαiy 2 A = +(1+ |y| )α. (E2.2) α 1+ |y|2

6 Here h·, ·i denotes the Euclidean inner product corresponding to the given quaternionic structure. Namely, for 2 vectors ξ and η we have hξ, ηi = Re(ξη¯) where in the right- hand side ξ and η are considered as quaternions. To express Aα it terms of y only we need to ﬁnd an expression for hy, Aαi. For that, take the inner product of both sides of equation (E2.2) with y. This leads to a linear equation for hy, Aαi. Solving this equation, we obtain (1 + |y|2)2 hy, A i = hy, αi. (E2.3) α 1 − |y|2

Substitute (E2.3) into (E2.2). We now have a formula for Aα in terms of y:

2 2hy, αiy Aα = (1+ |y| ) + α . (E2.4) 1 − |y|2

Diﬀerentiating equation (E2.4) along α we can see that

2 1+ |y| yAα Bα =2 2 hy, αi + . 1 − |y|2 1 − |y|2

Diﬀerentiate this equation once again along α. Then we ﬁnd that

1 6|A |2 ∂ B − B2 = α . α α 2 α (1 − |y|2)2

From equation (6) it follows that 6 C = (1 − |y|2)2 in this example. In particular, C may be nonzero. Note that in this example C is real everywhere.

4 Second Integrability Condition

The case C = 0 has been already worked out in the previous section. Now assume that C 6= 0 at a given point. A priori, C is a quaternion-valued function. But now we are going to prove the following

Lemma 4 The function C is real-valued.

This lemma does not follow form equation (1). We need to use directly the fact that f takes lines to circles. Proof. Take an arbitrary point x0 ∈ U and a vector α at x0. The line l passing through x0 in the direction of α, is mapped to a circle under f. Let t 7→ x = x(t) be the aﬃne parameterization of l such that x(0) = x0 andx ˙(0) = α. First assume that Bα is real at x0. Then f(l) must be a line. Therefore, Im(Bα) must vanish everywhere on l. It follows that C is also real in this case.

7 Suppose now that Im(Bα) 6= 0. By Proposition 5.4 from [5] the center of the circle f(l) is −1 f(x) − (ImBα(x)) Aα(x). Hence this expression does not depend on the choice of x ∈ l. Diﬀerentiate it by t:

−1 1 2 2 −1 −1 Aα + (ImBα) Im(B ) + Im(C)|Aα| (ImBα) Aα − (ImBα) BαAα =0. 2 α It follows that Im(C) = 0. The first integrability condition is a partial differential equation on B. Let us try to write down the integrability condition for this equation. We will use the language of quaternion-valued differential forms. By definition, the algebra of quaternion-valued differential forms is obtained from the algebra of ordinary real-valued differential forms by taking the tensor product with the quaternions over reals. In particular, this definition says how to multiply quaternion-valued forms. As was already mentioned, A and B are thought of as quaternion-valued 1-forms. From the first integrability condition it follows that 1 1 dB = B ∧ B + C(A ∧ A¯). (6′′′) 2 3 Take the differential of the both parts of this equation, using relations d2B = 0 and dA = 0: C dC ∧ A ∧ A¯ = (B ∧ A ∧ A¯ − A ∧ A¯ ∧ B). (7) 2 Proposition 5 Suppose that the values of the 1-forms A and B and the function C are given at some point, and the map α 7→ Aα is one-to-one at this point. If an analytic solution f of equations (2) with these initial values of A, B and C exists, then it is unique. Proof. Since A is non-degenerate, the operator of the right wedge multiplication by A ∧ A¯ is invertible. This can be verified by a simple direct computation. Therefore, equation (7) expresses all first derivatives of C through A and B. Consider this equation together with equations (2) and (6′′). We obtain a system of partial differential equations which expresses all first derivatives of A, B and C in terms of the values of A, B and C only. Therefore, by successive applications of this system we can express all higher derivatives of A, B and C at the given point through the initial values at this point. The proposition now follows. Note also that A, B and C determine the 3-jet of f. Namely, from equations (2) and (6′′) it follows that 3 2 2 Bx−x0 Ax−x0 ( 2 Bx−x0 + C|Ax−x0 | )Ax−x0 f = f(x0)+ A − + + + ... x x0 2 6 In particular, Proposition 5 can be reformulated as follows: Proposition 6 If there exists a diffeomorphism f : U ⊆ R4 → V ⊆ R4 with a given 3-jet at some point x0 ∈ U such that the image of each line segment lying in U is an arc of a circle or a line segment in V , then such diffeomorphism is unique.

8 5 Admissible 3-jets

From now on we identify the space of preimage with the space of image and assume that the distinguished point is the origin, f(0) = 0 and Aα = α for all vectors α at 0. This may be achieved by a linear change of variables in the preimage composed with a translation in the image. Thus if x = x0 + ix1 + jx2 + kx3 denotes the natural quaternionic coordinate, then A = dx = dx0 + idx1 + jdx2 + kdx3. Equation (7) imposes a restriction on B due to the fact that C is real. Let us ﬁnd all admissible values of B at 0 by solving (7) as a linear system on B and dC/C. We have A ∧ A¯ = dx ∧ dx = iω1 + jω2 + kω3 where ω1, ω2 and ω3 are real 2-forms given in coordinates by 1 0 3 2 ω1 = 2(dx ∧ dx + dx ∧ dx ), 2 0 1 3 ω2 = 2(dx ∧ dx + dx ∧ dx ), 3 0 2 1 ω3 = 2(dx ∧ dx + dx ∧ dx ). Denote by γ the real 1-form 2dC/C and suppose that B = B0 + iB1 + jB2 + kB3 where B0, B1, B2 and B3 are real 1-forms. Equation (7) can be now rewritten as the following system: 2 3 2(B ∧ ω3 − B ∧ ω2)= γ ∧ ω1, 3 1 2(B ∧ ω1 − B ∧ ω3)= γ ∧ ω2, (8) 1 2 2(B ∧ ω2 − B ∧ ω1)= γ ∧ ω3. µ µ Denote by Bν and γν (ν =0, 1, 2, 3) the components of the 1-forms B and γ so that

3 3 µ µ ν ν B = Bν dx , γ = γν dx . νX=0 νX=0 System (8) yields the following equations:

2 3 2 3 2 3 2 3 2(−B3 + B2 )= −γ1, 2(−B2 − B3 )= γ0, 2(B1 − B0 )= −γ3, 2(B0 + B1 )= γ2, 3 1 3 1 3 1 3 1 2(−B1 + B3 )= −γ2, 2(B0 + B2 )= γ3, 2(−B3 − B1 )= γ0, 2(B2 − B0 )= −γ1, 1 2 1 2 1 2 1 2 2(−B2 + B1 )= −γ3, 2(B3 − B0 )= −γ2, 2(B0 + B3 )= γ1, 2(−B1 − B2 )= γ0.

Exclude γν from these equations: 1 2 3 B1 = B2 = B3 , 1 2 3 B0 = B3 = −B2 , 2 3 1 B0 = B1 = −B3 , 3 1 2 B0 = B2 = −B1 .

The relations displayed above mean that Bα = p(α)+ αq where p is a real-valued 1-form and q is a quaternion. Hence the 3-jet at 0 of f should be as follows:

3 2 2 (p(x)+ xq)x ( (p(x)+ xq) + C|x| )x x + + 2 . (9) 2 6 Proposition 7 For an arbitrary real-valued linear function p on R4 and an arbitrary quaternion q, a local diﬀeomorphism with 3-jet (9) exists. Moreover, it can be chosen to be one of the classical projections.

9 Proof. First consider a map given by the implicit equation y x = C 2 . 1 − 6 |y|

C 2 This is clearly a map from example 2. It has the 3-jet x + 6 |x| x. This is a partial case of (9) where p = q = 0. To achieve other given values of p and q, compose this map with the following M¨obius transformation:

1 −1 −1 2 qy −1 y 7→ 2q 1 − 1 − 2q . 1 − 2 p(y) It is easy to see that the composition has 3-jet (9).

Concluding remarks. We should now make things add up in the proof of theorem 1. We have always assumed that ∂αAα = BαAα the multiplication by Bα being from the left. The case of the right multiplication is completely analogous. The only difference is that for C = 0 we would have the right quaternionic Hopf fibration instead of the left one. In general there is an open subset U ′ of U such that the multiplication ′ ′ by Bα is either form the left everywhere on U or from the right everywhere on U . Therefore theorem 1 holds on U ′. By the uniqueness theorem for analytic functions it holds then on U. If C = 0 everywhere on U, then by Proposition 3 we have a quaternionic Hopf fibration. Otherwise C nowhere vanishes on some open subset of U. Then by Propo- sitions 6 and 7 the map f is a classical projection on this subset. By the uniqueness theorem f is a classical projection on U.

Department of Mathematics, University of Toronto 100 St. George street, Toronto ON M5S3G3 Canada E-mail address: [email protected]

References

[1] K.Y. Lam: Some new results in composition of quadratic forms, Invent. Math. 79 (1985), 467-474 [2] G.S. Khovanskii: Foundations of Nomography, “Nauka”, Moscow, 1976 (Russian) [3] A.G. Khovanskii: Rectification of circles, Sib. Mat. Zh., 21 (1980), 221–226 [4] F.A. Izadi: Rectification of circles, spheres, and classical geometries, PhD thesis, University of Toronto, (2001) http://www.arxiv.org/abs/math.DG/0301220 [5] V.A. Timorin Rectification of Circles and Quaternions, Michigan Math. J. 51 (2003)